\[x-y\]
terms by completing the square and transforming variables.Example: Solve
\[x^2+2xy-2y^2=1\]
.Completing the square gives
\[(x+y)^2-3y^2=1\]
. Let \[z=x+y\]
then \[z^2-3y^2=1\]
.Convergents of the continued fraction representing
\[\sqrt{3}= [ 1, \lt 1,2 \gt ]\]
are\[\frac{1}{1}\]
\[1+ \frac{1}{1}=2\]
\[1+\frac{1}{1+ \frac{1}{2}}=\frac{5}{3}\]
\[1+\frac{1}{1+ \frac{1}{2+\frac{1}{1}}}=\frac{7}{4}\]
\[1+\frac{1}{1+ \frac{1}{2+\frac{1}{1+\frac{1}{2}}}}=\frac{19}{11}\]
\[1+\frac{1}{1+ \frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1}}}}}=\frac{26}{15}\]
Solutions are given by then convergent for the penultimate element of the cycle part of the continued fraction - 1. These are the convergents
\[\frac{2}{1}, \; \frac{7}{4}, \; \frac{26}{15}\]
.\[\frac{2}{1} z=x+y=2, \; y=1 \rightarrow x=1, \; y=1\]
\[\frac{7}{4} z=x+y=7, \; y=4 \rightarrow x=3, \; y=4\]
\[\frac{26}{15} z=x+y=26, \; y=15 \rightarrow x=11, \; y=15\]